Three-phase brushless DC motors have many uses, among which are as spindle motors for computer hard disk drivers, digital video disk (DVD) drivers, CD players, and tape-drives for video recorders. Such motors are recognized as having the highest torque and power capability for a given size and weight Compared to DC motors employing brushes, brushless DC motors enjoy reduced noise generation and improved reliability because no brushes need to be replaced due to wear.
FIG. 1 shows such a three-phase brushless DC motor 10 with three phases A, B, C having three coils 12, 14, 16 connected to each other in a Y-configuration at a center tap 18. As is well-known, the coils 12, 14, 16 are part of a stator that causes a permanent magnet rotor to rotate. The first coil 12 (phase A) is connected to a supply voltage Vret by a first high-side transistor 20 and to ground via a first low-side transistor 22 and a sense resistor 23; the second coil 14 (phase B) is connected to the supply voltage Vret by a second high-side transistor 24 and to ground via a second low-side ransistor 26 and the sense resistor; and the third coil 16 (phase C) is connected to the supply voltage Vret by a third high-side transistor 28 and to ground by a third low-side transistor 30 and the sense resistor 23. Each of the transistors is an NMOS transistor as is typical. Represented in FIG. 1 by voltage supply symbols are respective back EMF sources EA, EB, EC that are inherently induced by the permanent magnets of the rotor while the rotor is rotated.
This type of motor is driven by exciting its phases in a suitable sequence while always keeping two phases under power and leaving a third phase in tristate or floating with a high impedance (Z). For example, assume that initially the fist high-side transistor 20 and the second low-side transistor 26 are activated with high control signals on their gates while the other transistors are inactive. This results in a current IA through the first phase A having a value of +I, a current IB through the second phase having a value of -I, and zero current IC through the third phase as shown in FIG. 2. At predetermined instances (t1, t2, . . . ) the driving of the phase switches so that current is driven through the phase that was previously floatin and one of the other phases is left floating such that the algebraic sum of the currents in the three phases are always equal to zero. In FIG. 2 the driving sequence is as follows where the first letter indicates the phase of positive current flow and the second letter indicates the phase of negative current flow: EQU AB-AC-BC-BA-CA-CB.
In the instant of commutation from one stage to another (instances t1, t2, . . . ), if the current front were infinite, one would ideally find a system without perturbations. For example, at the instant t1, the phase A would maintain the current +I while the phases B and C would exchange the current flow, one from -I to 0 and the other from 0 to -I.
In reality, because of the presence of different time constants in the circuit, the commutation fronts of the two currents (IB and IC in the example of instant t1) would be non-ideal and non-synchronous. That is, the current IC increases more slowly than the current IB decreases. This translates into a variation of the current IA instead of the current IA remaining constant. The current variation generates torque ripple in the motor and much acoustic noise.
Analyzing the scheme of FIG. 1, it is possible to determine the reasons for the different commutation times in the two interested phases. At instant t1 (before the commutation) we would have:
VoutA=Vret IA=I PA1 VoutB=0 IB=-I PA1 VoutC=Vct IC=0 PA1 Vct=1/2 Vret PA1 VoutA=Vret PA1 VoutB=Vret+Vbe (due to the current of coil B recirculating in the intrinsic diode of second high-side transistor 24, where Vbe equals the drop across that intrinsic diode) PA1 VoutC=0 PA1 Vct=2/3 Vret. PA1 while the voltage across the third coil 16 is: EQU VoutC-(Vct+EC=0-(2/3Vret-E/2)=E/2-2/3Vret.
Given that the phases are out of phase by 120.degree., the electromotive forces driven will be instantaneously algebraically summed to zero. The commutation moreover happens at the instant t1 at the end of optimizing the torque ripple in the system. The back EMF in the three phases would have the following values: EA=E, EB=EC=-E/2, where E equals the maximum back EMF.
In the instant just after the commutation we would have:
The back EMF values will remain instantaneously unchanged.
The voltage across the second coil 14 is therefore: EQU VoutB-(Vct+EB)=Vret+Vbe-(2/3Vret-E/2)=1/3Vret+Vbe+E/2,
The two voltages will therefore be significantly different, creating different time constants of charge/discharge of the two currents (IB will be reduced more quickly than C will be increased).